Solving second order partial differential equation

Question

I' trying to solve this differential equation:

$$y^2 \frac{\partial ^2 u}{\partial y^2} - 2xy \frac{\partial ^2 u}{\partial x \partial y} + x^2 \frac{\partial ^2 u}{\partial x^2} + 2y \frac{\partial u}{\partial y} = 0$$

I've done following substitution: $$s = xy, \ \ t= y , \ \ u(x,y) = v(s, t)$$

and as a result I have $$t ^2 v _{t t} + 2s v_s + t v_t = 0$$ which I do not know how to solve.

Could you help me?

Thank you.

Answer 1

Indeed, $ r=\ln y,\;s=\ln x $ results in $ (\partial _{r}-\partial _{s})(\partial _{r}-\partial _{s}+1)u=0 $ One obvious solution is $ u=u(s+r) $. I leave it you to find the second.